The product of uncountably many copies of the real line is not a $k$-space For the purpose of this question, a $k$-space (or compactly generated space) is a space $X$ whose topology coincides with the final topology with respect to all the inclusions from its compact subspaces.  Equivalently, a set $U\subseteq X$ is open iff the intersection $U\cap C$ is open in $C$ for every compact subspace $C$ of $X$.
Willard, exercise 43H (part 2) says:

The product of uncountably many copies of $\mathbb R$ is not a $k$-space.

So let $I$ be some uncountable index set.  The set $\omega=\{0,1,2,...\}$ is closed in $\mathbb R$, so $\omega^I$ is closed in $\mathbb R^I$.  Since any closed subspace of $k$-space is a $k$-space, it is sufficient to show:

Give $\omega$ the discrete topology.  The product of uncountably many copies of $\omega$ is not a $k$-space.

Proof (incomplete): We have to show that $X=\omega^I$ with the product topology is not a $k$-space.  Following the hint in Willard, take $T=\bigcup_{n=1}^\infty T_n\subseteq X$ where $T_n$ is the set of points $x$ such that $x_i=0$ for at most $n$ coordinates and $x_i=n$ for all the other coordinates.  The claim is that $T$ is not closed, but is $k$-closed.
(1) $T$ is not closed: The point $\mathbf{0}\in X$ with all coordinate equal to $0$ is in $\overline T$, because every basic nbhd of it consists of all points that are zero on some finite set $J\subseteq I$, and that will contain some point in $T_n$ for $n$ large enough.
No other point outside of $T$ is in $\overline T$.  That is easy to check.  For example if a point $x$ has at least two non-zero values $x_i=n$ and $x_j=m$, the nbhd of $x$ defined by these conditions is disjoint from $T$.  And similarly if almost all values $x_i$ are equal to $n$, but $x$ has more than $n$ zeros.  So $\overline T=T\cup\{\mathbf 0\}$.
(2) $T$ is $k$-closed:  We have to show that given any compact $C\subseteq X$, the set $T\cap C$ is closed in $C$.
How to show this last part?
 A: Borrowing some ideas from R. Brown, Ten topologies for $X\times Y$,
Quarterly J. Math., 14(1) (1963), pp. 303-319, here is a solution for part (2).
Given a compact set $C$ in $X$, we have to show that $T\cap C$ is closed in $C$.  Since $X$ is Hausdorff, $C$ is closed in $X$ and it is equivalent to show that $T\cap C$ is closed in $X$.
For each $i\in I$, the projection $\pi_i(C)$ of $C$ onto the $i$-th component is compact, so there is some integer $n$ such that $\pi_i(C)\subseteq [0,n]$.  For each $n$, let $I_n=\{i\in I:\pi_i(C)\subseteq [0,n]\}$. Then $I$ is the countable union of the $I_n$ for $n=1,2,\dots$.  Since $I$ is uncountable, at least one these sets, say $I_m$, is infinite. (One could even choose an uncountable one, but an infinite one is good enough.)  Now, as shown in part (1), $\overline T=T\cup\{\mathbf 0\}$. So to show that $T\cap C$ is closed in $X$, it is enough to find a nbhd of $\mathbf 0$ that is disjoint from $T\cap C$.  Since $I_m$ is infinite, we can choose a subset $J\subseteq I_m$ with $m+1$ elements.  The nbhd of $\mathbf 0$ consisting of those point that are $0$ at elements of $J$ is disjoint from $T\cap C$ by construction.  This completes the proof.
